Then the triangle ABC shall be equal to the triangle DBC. Construction. Produce AD both ways to the points E, F; through B draw BE, parallel to CA (I. 31); and through C draw CF parallel to BD. Demonstration. Then and 1. Each of the figures EBCA, DBCF is a parallelogram; 2. EBCA is equal to DBCF (I. 35), because they are upon the same base BC, and between the same parallels BC, EF. And because the diameter AB bisects the parallelogram EBCA, therefore 3. The triangle ABC is half of the parallelogram EBCA (I. 34). Also, because the diameter DC bisects the parallelogram DBCF, therefore 4. The triangle DBC is half of the parallelogram DBCF, but the halves of equal things are equal (Ax. 7); therefore 5. The triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q.E.D. PROPOSITION 38.-Theorem. Triangles upon equal bases and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD. Then the triangle ABC shall be equal to the triangle DEF. Construction. Produce AD both ways to the points G, H; through B draw BG parallel to CA (I. 31); and through F draw FÍ parallel to ED. A H B Demonstration. Then and 1. Each of the figures GBCA, DEFH is a parallelogram; 2. GBCA, DEFH are equal to one another (I. 36), because they are upon equal bases BC, EF, and between the same parallels BF, GH. And because the diameter AB bisects the parallelogram GBCA; therefore 3. The triangle ABC is the half of the parallelogram GBCA (I. 34); Also, because the diameter DF bisects the parallelogram DEFH, therefore 4. The triangle DEF is the half of the parallelogram DEFH, but the halves of equal things are equal (Ax. 7); therefore 5. The triangle ABC is equal to the triangle DEF. Wherefore, triangles upon equal bases, &c. Q.E.D. PROPOSITION 39.-Theorem. Equal triangles upon the same base and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it. Then the triangles ABC, DBC shall be between the same parallels. D B Construction. Join AD; then AD shall be parallel to BC. For if AD be not parallel to BC, if possible, through the point A, draw AE parallel to BC (I. 31), meeting BD, or BD produced, in E, and join EC. Demonstration. Then 1. The triangle ABC is equal to the triangle EBC (I. 37), because they are upon the same base BC, and between the same parallels BC, AE; but the triangle ABC is equal to the triangle DBC (hyp.); therefore 2. The triangle DBC is equal to the triangle EBC, the greater triangle equal to the less, which is impossible; therefore 3. AE is not parallel to BC. In the same manner it can be demonstrated, that no other line drawn from A but AD is parallel to BC; therefore 4. AD is parallel to BC. Wherefore, equal triangles upon, &c. Q.E.D. PROPOSITION 40.-Theorem. Equal triangles upon equal bases in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the same parts. Then the triangles ABC, DEF shall be between the same parallels. Construction. Join AD; then AD shall be parallel to BF. For, if AD be not parallel to BF, if possible, through A, draw AG parallel to BF (I. 31), meeting ED, or ED produced in G, and join GF. 1. The triangle ABC is equal to the triangle GEF (I. 38), because they are upon equal bases BC, EF, and between the same parallels BF, AG; but the triangle ABC is equal to the triangle DEF (hyp.); therefore 2. The triangle DEF is equal to the triangle GEF (Ax. 1), the greater triangle equal to the less, which is impossible; therefore 3. AG is not parallel to BF. And in the same manner it can be demonstrated, that there is no other line drawn from A parallel to it but AD; therefore 4. AD is parallel to BF. Wherefore, equal triangles upon, &c. Q.E.D. PROPOSITION 41.-Theorem. If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle. Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE. Then the parallelogram ABCD shall be double of the triangle EBC. D A B Construction. Join AC. Demonstration. Then D B 1. The triangle ABC is equal to the triangle EBC (I. 37), because they are upon the same base BC, and between the same parallels BC, AE. But 2. The parallelogram ABCD is double of the triangle ABC, because the diameter AC bisects it (I. 34); wherefore also 3. ABCD is double of the triangle EBC. Therefore, if a parallelogram and a triangle, &c. Q.E.D. PROPOSITION 42.-Problem. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Construction. Bisect BC in E (I. 10), and join AE; at the point E, in the straight line EC, make the angle CEF equal to the angle D (I. 23); through C draw CG parallel to EF; and through A draw AFG parallel to BC (I. 31), meeting EF in F, and CG in G. |